How many three-digit numbers are divisible by | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The first three-digit number divisible by $7$ is $105$.The next numbers are $105 + 7 = 112, 112 + 7 = 119, \ldots$These numbers form an Arithmetic

Vedclass · Accepted Answer

$128$

Vedclass · Answer

(C) The first three-digit number divisible by $7$ is $105$.The next numbers are $105 + 7 = 112, 112 + 7 = 119, \ldots$These numbers form an Arithmetic Progression $(A.P.)$ where the first term $a = 105$ and the common difference $d = 7$.The largest three-digit number is $999$. Dividing $999$ by $7$,we get $999 = 7 	imes 142 + 5$. The remainder is $5$.Thus,the largest three-digit number divisible by $7$ is $999 - 5 = 994$.Let $994$ be the $n^{th}$ term of the $A.P.$ sequence.Using the formula $a_n = a + (n - 1)d$:$994 = 105 + (n - 1)7$$994 - 105 = (n - 1)7$$889 = (n - 1)7$$n - 1 = 889 / 7$$n - 1 = 127$$n = 128$Therefore,there are $128$ three-digit numbers divisible by $7$.

How many three-digit numbers are divisible by $7$?

Similar Questions

Show that $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ form an $AP$ where $a_{n}$ is defined as $a_{n}=3+4 n$. Also,find the sum of the first $15$ terms.

$2, \frac{5}{2}, 3, \frac{7}{2}, \ldots$ are $APs$? If they form an $AP$,find the common difference $d$ and write three more terms.

Find the sum of the following $APs$ $-37, -33, -29, \ldots$ to $12$ terms.

Find the sum of the following $APs$: $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots,$ to $11$ terms.

Vedclass Test Series

Exam Paper Generator

Online Exam Module